Find the component of gravity along the curve

  • #1
Homework Statement:
I only need help with how to find the component of gravity along the curve.

Here is the problem statement:

A frictionless surface is in the shape of a function which has its endpoints at the same height but is otherwise arbitrary. A chain of uniform mass per unit length rests on this surface (from end to end; see Fig. 1.10). Show that the chain will not move.
Relevant Equations:
None.
I only need help with how to find the component of gravity along the curve. See image.
 

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Answers and Replies

  • #2
haruspex
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how to find the component of gravity along the curve.
Seems too easy... call the angle the slope makes to the horizontal θ. What is the component in terms of that? How does that angle relate to dy/dx?
 
  • #3
I did, but then I always get a relationship with theta and the problem should not include theta in the expression.
 
  • #4
haruspex
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I did, but then I always get a relationship with theta and the problem should not include theta in the expression.
That's why I asked the second question in post #2.
 
  • #5
I thought about it but couldn't make a connection to make progress. tan(theta) = dy/dx
 
  • #6
haruspex
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I thought about it but couldn't make a connection to make progress. tan(theta) = dy/dx
Right, and you are using f' for dy/dx.
What expression did you get for the component of mg in terms of θ?
 
  • #7
That's why I asked the second question in post #2.

Please see attached, c is the component of gravity along the curve (hypotenuse). Have I done anything wrong?
 

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  • #8
haruspex
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Please see attached, c is the component of gravity along the curve (hypotenuse). Have I done anything wrong?
I do not understand where some of your equations come from. It doesn't help that the order on the page doesn’t seem to match the order of your reasoning.
Please stop posting images of your working, it is against forum rules.
Take the trouble to type it in, in logical order. Then I will be able to make reference to specific steps more easily.
 
  • #9
$$ T
I do not understand where some of your equations come from. It doesn't help that the order on the page doesn’t seem to match the order of your reasoning.
Please stop posting images of your working, it is against forum rules.
Take the trouble to type it in, in logical order. Then I will be able to make reference to specific steps more easily.

My apology, let's start from the beginning.

Consider an infinitesimal piece of the chain between x and x+dx. The opposite side is given by f'dx.

The hypotenuse of the chain can be found by Pythagorean's theorem.

$$ c^2 = dx^2 + f'^2 dx^2 = dx^2 (1+f'^2)$$
Then ##c = dx \sqrt{1+f'^2}##

Let's call the density (mass per unit length) of the chain ##\rho##, then the mass of this piece is given by ##\rho \sqrt{1+f'^2}##

The question is, what is the component of gravity along the curve?

First, the force of gravity gravity is given by ## mg = \rho g \sqrt{1+f'^2} dx ##. In a right triangle, that would be the opposite side.

We also know that ## tan \theta = f' ##, using that the tangent of an angle is opposite over adjacent, we get
$$ f' = \rho g \frac{\sqrt{1+f'^2} dx}{a}$$
Where a is the adjacent portion of the triangle representing the horizontal component of gravity.

Solving for the adjacent side, we get ## a = \rho g \frac{ \sqrt{1+f'^2}dx }{f'} ##

Then, in order to find the hypotenuse (the component of gravity along the chain, I use Pythagorean's theorem:
$$c^2 =a^2 + b^2$$
$$c^2 = (\rho g \frac{\sqrt{1+f'^2} dx }{f'})^2 + (\rho g \sqrt{1+f'^2} dx )^2$$
Then, getting the common denominator:
$$c^2 = \rho^2 g^2 \frac{1+f'^2 dx^2 }{f'^2} + \rho^2 g^2 \frac{(1+f'^2) dx^2 f'^2}{f'^2}$$
Then factoring out some terms
$$c^2 = \rho^2 g^2 \frac{1+f'^2 dx^2 }{f'^2} (1+f'^2)$$
Simplifying..
$$c^2 = \rho^2 g^2 \frac{(1+f'^2)^2 dx^2 }{f'^2} $$
Then square rooting we get:
$$c = \rho g \frac{(1+f'^2) dx }{f'} $$
Which is the portion of gravity in the direction of the curve of the chain.
 
  • #10
haruspex
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we get
$$ f' = \rho g \frac{\sqrt{1+f'^2} dx}{a}$$
Where a is the adjacent portion of the triangle representing the horizontal component of gravity.
Gravity does not have a horizontal component, the normal force does.
To avoid getting tangled up with the normal force, write the force balance equation parallel to the slope.
 
  • #11
Gravity does not have a horizontal component, the normal force does.
To avoid getting tangled up with the normal force, write the force balance equation parallel to the slope.

There is ##mg sin \theta## pulling it to the left and the frictional force holding it to right parallel to the slope with a magnitude ##\mu N##

So the sum of the forces along the slope would be
$$\mu N - mg~sin\theta = 0$$
Upon substituting the ##m=\rho \sqrt{1+f'^2}dx## we get:
$$\mu N - \rho g \sqrt{1"f'^2}dx ~sin\theta=0$$
 
  • #12
ehild
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There is ##mg sin \theta## pulling it to the left and the frictional force holding it to right parallel to the slope with a magnitude ##\mu N##

So the sum of the forces along the slope would be
$$\mu N - mg~sin\theta = 0$$
Upon substituting the ##m=\rho \sqrt{1+f'^2}dx## we get:
$$\mu N - \rho g \sqrt{1"f'^2}dx ~sin\theta=0$$
The surface is frictionless. Write sin(theta) in terms of tan(theta) . The whole force that acts on the chain should be zero.
 
  • #13
The surface is frictionless. Write sin(theta) in terms of tan(theta) . And you need the whole force that acts on the chain.

##sin\theta = tan \theta cos \theta. ## Can you clarify what you mean by the last sentence? As far as Ia m concerned the force acting on the chain is just the weight of the chain, but I don't know how to incorporate that.
 
  • #14
ehild
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##sin\theta = tan \theta cos \theta. ## Can you clarify what you mean by the last sentence? As far as Ia m concerned the force acting on the chain is just the weight of the chain, but I don't know how to incorporate that.
You know that sin(θ2)+cos(θ2)=1. You can express sin(θ) by tan(θ).
The piece of the chain is attached to the neighbor pieces, and they act with some force on the piece.
So the force of gravity along the slope is not zero. The net force is.
You do not know these forces of interaction. But in order the chain be motionless, the whole force on the chain should be zero. Adding all forces, the forces of interaction will cancel.
 
  • #15
You know that sin(θ2)+cos(θ2)=1. You can express sin(θ) by tan(θ).
The piece of the chain is attached to the neighbor pieces, and they act with some force on the piece.
So the force of gravity along the slope is not zero. The net force is.
You do not know these forces of interaction. But in order the chain be motionless, the whole force on the chain should be zero. Adding all forces, the forces of interaction will cancel.

I'll call the normal force on one piece of the chain attached to the neighboring piece ## N_{ConC}##.

Summing the forces in parallel slope direction, one gets ## N_{ConC} - mgsin\theta = 0##

Meanwhile, using what you told me:
$$sin^\theta + cos^2\theta = 1$$
$$tan^2\theta~ cos^2\theta + cos^2 ~\theta = 1$$

Not sure what to plug in at that point, I know ##tan^2 \theta = f'^2 ##, but I feel that I shouldn't be mixing it up with the component of gravity parallel to the slope.
 
  • #16
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I have a different approach to this. Let s be the arc length along the chain and surface, and let ##delta## represent a differential segment along this arc. The mass of chain within ##\Delta s## is ##\rho \Delta s## and its weight is ##\rho g \Delta s##. A force balance on this segment (assuming no acceleration) is $$-\rho g \mathbf{i_y}\Delta s+[T\mathbf{i_t}]_{s+\Delta s}-[T\mathbf{i_t}]_{s}+N\mathbf{i_n}\Delta s=0\tag{1}$$where T is the tension in the chain, N is the normal force per unit length exerted by the surface on the chain, ##\mathbf{i_y}## is the unit vector in the vertical direction, ##\mathbf{i_t}## is the unit tangent vector to the surface, and ##\mathbf{i_n}## is the unit normal vector to the surface, with:
$$\mathbf{i_t}=\cos{\phi}\mathbf{i_x}+\sin{\phi}\mathbf{i_y}\tag{2}$$
$$\mathbf{i_n}=-\sin{\phi}\mathbf{i_x}+\cos{\phi}\mathbf{i_y}\tag{3}$$
and$$\cos{\phi}=\frac{dx}{ds}\tag{4}$$
$$\sin{\phi}=\frac{dy}{ds}\tag{5}$$
If we divide Eqn. 1 by ##\Delta s## and take the limit as ##\Delta s## approaches zero, we obtain:
$$-\rho g \mathbf{i_y}+\frac{d(T\mathbf{i_t})}{ds}+N\mathbf{i_n}=0\tag{6}$$Substitution of Eqns. 2 and 3 into Eqns. 6 then yields $$-\rho g \mathbf{i_y}+\frac{dT}{ds}\mathbf{i_t}+T\frac{d\phi}{ds}\mathbf{i_n}+N\mathbf{i_n}=0\tag{7}$$If we take the dot product of this equation with the unit vector in the tangent direction, we obtain:$$\frac{dT}{ds}=\rho g \sin{\phi}=\rho g \frac{dy}{ds}\tag{8}$$From this, if follows that $$\Delta T=\rho g\Delta y$$
 
  • #17
To Chestermiller, thank you, but the problem is asking to show that the total force along the curve is 0. You did help me out though in writing the sum of the forces along the slope as ehild asked.

To ehild: The sum of the forces along the slope is given by"

$$ \sum F_{Tangential} = -\rho g \Delta s~sin\theta + T_{s+ \Delta s} - T_s = 0$$

Also, keep in mind, I still want to know how to get the tangential component of the gravity along the slope as the book did. I know that force of gravity is given by ##-g f' \sqrt{1+f'^2} ## But I can't follow why the component of gravity along the curve is supposed to be given by ## \frac{-gf'}{\sqrt{1+f'^2}}##.
 
  • #18
ehild
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I'll call the normal force on one piece of the chain attached to the neighboring piece ## N_{ConC}##.

Summing the forces in parallel slope direction, one gets ## N_{ConC} - mgsin\theta = 0##

Meanwhile, using what you told me:
$$sin^\theta + cos^2\theta = 1$$
$$tan^2\theta~ cos^2\theta + cos^2 ~\theta = 1$$
Solve the equation for cos2(θ):
##\cos^2(\theta)=\frac{1}{1+\tan^2(\theta)}##
so
##\sin(\theta)=\frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}}##

Not sure what to plug in at that point, I know ##tan^2 \theta = f'^2 ##, but I feel that I shouldn't be mixing it up with the component of gravity parallel to the slope.
Substitute f' for tan(theta), and use the new expression for sin(theta) in the formula for the tangential component of gravity.
 
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  • #19
ehild
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To ehild: The sum of the forces along the slope is given by"

$$ \sum F_{Tangential} = -\rho g \Delta s~sin\theta + T_{s+ \Delta s} - T_s = 0$$
whole
This is the tangential force on a piece of the chain. You need the tangential force exerted on the whole chain.
 
  • #20
haruspex
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To Chestermiller, thank you, but the problem is asking to show that the total force along the curve is 0. You did help me out though in writing the sum of the forces along the slope as ehild asked.

To ehild: The sum of the forces along the slope is given by"

$$ \sum F_{Tangential} = -\rho g \Delta s~sin\theta + T_{s+ \Delta s} - T_s = 0$$

Also, keep in mind, I still want to know how to get the tangential component of the gravity along the slope as the book did. I know that force of gravity is given by ##-g f' \sqrt{1+f'^2} ## But I can't follow why the component of gravity along the curve is supposed to be given by ## \frac{-gf'}{\sqrt{1+f'^2}}##.
It is a lot simpler to work in terms of trig functions of the angle of the slope instead of f' and square roots.
Given an element that spans a vertical height of dy, what is its weight?
What is the angle between that vertical vector and the slope? So what its tangential component?
For each such element on a left-facing slope there must be a corresponding element on a right-facing slope.
 
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  • #21
Solve the equation for cos2(θ):
##\cos^2(\theta)=\frac{1}{1+\tan^2(\theta)}##
so
##\sin(\theta)=\frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}}##


Substitute f' for tan(theta), and use the new expression for sin(theta) in the formula for the tangential component of gravity.
Oh, got it! Thank you!
 

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